HOMEWORK 8 FOR 18.745, FALL 2012
DUE FRIDAY, NOVEMBER 2 BY 3PM.
The base field is algebraically closed of characteristic zero and the Lie algebras
are finite dimensional unless stated otherwise.
(1) Let t be a maximal toral subalgebra in a semisimple Lie algebra g. Let
Σ = Σ(g) = {α ∈ t∗ | α 6= 0 &gα 6= 0}, so that, as has been shown in class,
Σ ⊂ E := Q · Σ ⊗Q R is a root system, where the form on E comes from
the Killing form on g.
(a) Recall that for every α ∈ Σ we have an sl2 triple eα ∈ gα , fα ∈ g−α ,
hα ∈ t. Set Σ0 = {hα | α ∈ Σ}. Show that Σ0 ⊂ Q · Σ0 ⊗Q R ∼
= E ∗ is
also a root system.
(b) Let g = sp6 . The root system Σ0 is not isomorphic to Σ, but it is
isomorphic to Σ(g0 ) for another semi-simple Lie algebra g0 . Describe
g0 and prove this statement.
[Here we say that two irreducible root systems Σ1 ⊂ E1 , Σ2 ⊂ E2 are
isomorphic if there exists an isomorphism φ : E1 ∼
= E2 sending Σ1 to
Σ2 and satisfying (φ(x), φ(y)) = C(x, y) for some constant c].
[Hint: You may want to check that Σ0 is isomorphic to the root system
α
| α ∈ Σ}.
Σˇ= { (α,α)
(2) Check that for any nontrivial representation of sl2 the corresponding trace
form is nondegenerate.
(3) Let g be a semi-simple Lie algebra and x ∈ g be a semi-simple element.
Show that the centralizer of x is a reductive Lie algebra.
[Hint: Let t be a maximal toral subalgebra containing x. Then the
centralizer is the sum of t and gα for α such that α(x) = 0. Recall Cartan
criterion and use the previous problem to show that the kernel of the Killing
form of the centralizer is contained in the subspace of y ∈ t, s.t. α(y) = 0
for α as above.]
(4) (Optional) Let g be any Lie algebra.
(a) Show that the tensor product of two semi-simple g modules is semisimple.
(b) Let k be the intersection of kernels of all irreducible representations
of g. Show that g/k is reductive, and that it is a maximal reductive
quotient of g (i.e. for any ideal k0 in g such that g/k0 is reductive we
have: k ⊂ k0 ).
[Hint: use the optional problem from the previous problem set.]
(5) (Optional) Let g be a Lie algebra over R. Show that the Killing form of g
can not be positive definite.1
[Hint: Pick x ∈ g and consider the eigenvalues of ad(x) acting on g ⊗ C.
If there exists an eigenvalue with a nonzero real part deduce that g contains
and ad nilpotent element y, then κ(y, y) = 0. Otherwise all eigenvalues of
ad(x) are purely imaginary, so kappa(x, x) ≤ 0.]
1It may however be negative definite.
1